A ball of mass `m` moving with velocity `V`, makes a head on elastic collision with a ball of the same moving with velocity `2 V` towards it. Taking direction of `V` as positive velocities of the two balls after collision are.

Options

A. `- V and 2 V`
B. `2 V and - V`
C. `V and - 2 V`
D. `- 2 v and V`

Explanation

To solve the problem of the head-on elastic collision between two balls of equal mass, we can follow these steps:

### Step 1: Understand the Initial Conditions
We have two balls of mass `m`.
- Ball 1 is moving with velocity `V` (positive direction).
- Ball 2 is moving towards Ball 1 with velocity `2V` (negative direction).

### Step 2: Apply Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. The conservation of momentum can be expressed as:
\[ m \cdot V + m \cdot (-2V) = m \cdot v_1 + m \cdot v_2 \]
Where \( v_1 \) and \( v_2 \) are the velocities of Ball 1 and Ball 2 after the collision.

### Step 3: Simplify the Momentum Equation
The equation simplifies to:
\[ mV - 2mV = mv_1 + mv_2 \]
\[ -mV = mv_1 + mv_2 \]
Dividing through by `m` gives:
\[ -V = v_1 + v_2 \]
(1)

### Step 4: Apply Conservation of Kinetic Energy
The conservation of kinetic energy for an elastic collision can be expressed as:
\[ \frac{1}{2}mV^2 + \frac{1}{2}m(2V)^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \]
This simplifies to:
\[ \frac{1}{2}mV^2 + 2mV^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \]
\[ \frac{5}{2}mV^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \]
Dividing through by `m` and multiplying by 2 gives:
\[ 5V^2 = v_1^2 + v_2^2 \]
(2)

### Step 5: Solve the Equations
Now we have two equations:
1. \( -V = v_1 + v_2 \)
2. \( 5V^2 = v_1^2 + v_2^2 \)

From equation (1), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = -V - v_1 \]

Substituting this into equation (2):
\[ 5V^2 = v_1^2 + (-V - v_1)^2 \]
Expanding the second term:
\[ 5V^2 = v_1^2 + (V^2 + 2Vv_1 + v_1^2) \]
Combining like terms:
\[ 5V^2 = 2v_1^2 + 2Vv_1 + V^2 \]
Rearranging gives:
\[ 4V^2 = 2v_1^2 + 2Vv_1 \]
Dividing through by 2:
\[ 2V^2 = v_1^2 + Vv_1 \]
Rearranging this into standard quadratic form:
\[ v_1^2 + Vv_1 - 2V^2 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula:
\[ v_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = V, c = -2V^2 \):
\[ v_1 = \frac{-V \pm \sqrt{V^2 + 8V^2}}{2} \]
\[ v_1 = \frac{-V \pm 3V}{2} \]
This gives two possible solutions:
1. \( v_1 = V \) (discarded since it leads to a non-physical situation)
2. \( v_1 = -2V \)

### Step 7: Find \( v_2 \)
Using \( v_2 = -V - v_1 \):
\[ v_2 = -V - (-2V) = V \]

### Final Velocities
Thus, the velocities of the two balls after the collision are:
- Ball 1: \( v_1 = -2V \) (moving in the negative direction)
- Ball 2: \( v_2 = V \) (moving in the positive direction)

### Summary
The final velocities of the two balls after the collision are:
- Ball 1: \( -2V \)
- Ball 2: \( V \)